# Prove-that-x-y-R-7x-2-6xy-2y-2-x-3-gt-0-

Question Number 11834 by Mr Chheang Chantria last updated on 02/Apr/17
$$\boldsymbol{{Prove}}\:\boldsymbol{{that}}\:\forall\boldsymbol{{x}},\boldsymbol{{y}}\in\boldsymbol{{R}} \\$$$$\Rightarrow\mathrm{7}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{{xy}}+\mathrm{2}\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{x}}+\mathrm{3}\:>\:\mathrm{0} \\$$
Answered by mrW1 last updated on 02/Apr/17
$$\mathrm{7}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{{xy}}+\mathrm{2}\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{x}}+\mathrm{3} \\$$$$=\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}{x}\right)^{\mathrm{2}} −\mathrm{2}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}{x}×\sqrt{\mathrm{2}}{y}+\left(\sqrt{\mathrm{2}}{y}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}{x}\right)^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{1}+\left[\mathrm{7}−\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \right]{x}^{\mathrm{2}} +\left(\mathrm{3}−\mathrm{1}\right) \\$$$$=\left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}{x}−\sqrt{\mathrm{2}}{y}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{1}\right)^{\mathrm{2}} +\left(\frac{\mathrm{3}}{\mathrm{2}}{x}\right)^{\mathrm{2}} +\mathrm{2} \\$$$$>\mathrm{2}>\mathrm{0} \\$$
Commented by Mr Chheang Chantria last updated on 03/Apr/17
$$\left.\boldsymbol{{Good}}\:\boldsymbol{{solution}}\:;\right) \\$$$$\\$$
Answered by ajfour last updated on 02/Apr/17
$$\mathrm{6}{x}^{\mathrm{2}} −\mathrm{6}{xy}+\mathrm{2}{y}^{\mathrm{2}} +{x}^{\mathrm{2}} +{x}+\mathrm{3} \\$$$$=\mathrm{2}{x}^{\mathrm{2}} \left[\mathrm{3}−\frac{\mathrm{3}{y}}{{x}}+\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} \:\right]+\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{3} \\$$$$=\mathrm{2}{x}^{\mathrm{2}} \left[\left(\frac{{y}}{{x}}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{3}\:\right]+\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{4}} \\$$$$=\mathrm{2}{x}^{\mathrm{2}} \left[\frac{\left(\mathrm{2}{y}−\mathrm{3}{x}\right)^{\mathrm{2}} }{\mathrm{4}{x}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{4}}\right]+\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{4}} \\$$$$=\frac{\left(\mathrm{2}{y}−\mathrm{3}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{4}}\:>\mathrm{0} \\$$
Commented by Mr Chheang Chantria last updated on 03/Apr/17
$$\left.\boldsymbol{\mathrm{that}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{sweet}}\:;\right) \\$$$$\\$$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 02/Apr/17
$$\mathrm{7}{x}^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{6}{y}\right){x}+\mathrm{2}{y}^{\mathrm{2}} +\mathrm{3}=\mathrm{0} \\$$$$\Delta=\left(\mathrm{1}−\mathrm{6}{y}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{7}\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{3}\right)= \\$$$$\mathrm{1}−\mathrm{12}{y}+\mathrm{36}{y}^{\mathrm{2}} −\mathrm{56}{y}^{\mathrm{2}} −\mathrm{84}= \\$$$$−\mathrm{20}{y}^{\mathrm{2}} −\mathrm{12}{y}−\mathrm{83}=−\left(\mathrm{20}{y}^{\mathrm{2}} +\mathrm{12}{y}+\mathrm{83}\right) \\$$$$\Delta^{'} =\mathrm{12}^{\mathrm{2}} −\mathrm{4}×\mathrm{20}×\mathrm{83}<\mathrm{0} \\$$$${because}\:\Delta'<\mathrm{0},{then}\:\Delta\:{only}\:{have}\:{one} \\$$$${sign}\:{that}\:{similar}\:{to}\:{the}\:{coefficent}\:{of} \\$$$${y}^{\mathrm{2}\:} \:{i}.{e}\::\left(−\mathrm{20}\right).{so}\:{alwyes}\:\Delta<\mathrm{0}\:{and}\:{the} \\$$$$\mathrm{7}{x}^{\mathrm{2}} −\mathrm{6}{xy}+\mathrm{2}{y}^{\mathrm{2}} +\mathrm{3},{anywere}\:{have}\:{one} \\$$$${sign}\:{that}\:{similar}\:{to}\:{sign}\:{of}\:{x}^{\mathrm{2}} ,{i}.{e}:\left(+\mathrm{7}\right) \\$$$${so}\:{this}\:{polynomial}\:{is}\:{positive}\:{anywere}. \\$$
Commented by mrW1 last updated on 02/Apr/17
$${good}\:{and}\:{interesting}\:{point}\:{of}\:{view}! \\$$
Commented by Mr Chheang Chantria last updated on 03/Apr/17
$$\boldsymbol{{nice}}\:\boldsymbol{{solution}}\:\boldsymbol{{and}}\:\boldsymbol{{nice}}\:\boldsymbol{{explain}} \\$$$$\left.\boldsymbol{{Thanks}}\:\boldsymbol{{you}}\:;\right) \\$$